Solve ∫ 6^2+(sqrt{3})^2

$$\int 6 ^ { 2 } + ( \sqrt { 3 } ) ^ { 2 }$$

Evaluate

$$39x+С$$

Solution Steps

Calculate 6 to the power of 2 and get 36.

$$\int 36+\left(\sqrt{3}\right)^{2}\mathrm{d}x$$

The square of \sqrt{3} is 3.

$$\int 36+3\mathrm{d}x$$

Add 36 and 3 to get 39.

$$\int 39\mathrm{d}x$$

Find the integral of 39 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$39x$$

If F\left(x\right) is an antiderivative of f\left(x\right), then the set of all antiderivatives of f\left(x\right) is given by F\left(x\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$39x+С$$

Differentiate w.r.t. x

$$39$$